Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Section 1.2 - Mathematical Models: A Catalog of Essential Functions. - 1.2 Exercises: 16


(a) $d=48 t$ (b) the graph is on the figure below (c) the slope is $48$. It represents velocity in miles per hour.

Work Step by Step

The distance traveled $d$ will be a linear function of time: $$d=vt+d_0$$ where $v$ and $d_0$ are constants. (a) We have to find $v$ and $d_0$. When $t=0$ then we need that $d=0$ because Jason hasn't traveled any distance: $$0=v\cdot0+d_0\Rightarrow d_0=0.$$ Now we have $d=vt.$ To find $v$ note that for $50$ minutes which is $50/60=5/6$ hours Jason travels the distance of $40$ miles so we have $$40=v\cdot \frac{5}{6}\Rightarrow v=48.$$ This means that the distance traveled in terms of elapsed time is $$d=48t$$ where $d$ is in miles and $t$ is in hours. (b) The graphs contains the origin because when $t=0$ then $d=48\cdot0=0$. Further, when $t=1$ then $d=48\cdot 1=48$ so we just draw the line starting at the origin and passing through the point $(1,48)$. This graph is on the figure below. (c) The slope is the constant that multiplies the variable $t$ which is in this case $v=48$. It says the distance in miles that Jason travels for one hour i.e. the velocity in miles per hour.
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